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find the equation of the locus of a point that moves so that its distance from the line 3x-4y+9=0 is always 5

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let that point be P (x, y)
distance from a point to a line = \(\frac{Ax+By+C}{\sqrt{A^2 + B^2}}\)
can you put the equation for distance ? from point P(x, y) to line 3x-4y+9=0

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Other answers:

oh so it would be (3x-4y+9)/square root( 3^2 +(-4)^2)
(3x -4y+9)/5 =5 3x -4y+9 =25 3x-4y -16 =0
is that right ganeshi8?
btw i thought there was an ABSOLUTE SIGN around Ax + By + C for the formula
yep! u r right.. we must get two lines
|(3x -4y+9)|/5 =5
3x-4y+9 = 25, and, 3x-4y + 9 = -25
ngawww i get it !
it tricked me.... i was thinking there should be two lines at the very end. nice catch :)
thanks again - LOL and about writing GAPS yesterday, it wasn't meant to be rude or anything, just expressing my gratitude for your help ;)
lol fine i get it nw :))
two lines we just got... which are at a distance of 5 from the given line... they look as above

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